A noninequality for the fractional gradient

Abstract

In this paper we give a streamlined proof of an inequality recently obtained by the author: For every alpha is an element of (0, 1) there exists a constant C = C(alpha, d) > 0 such that parallel to u parallel to(Ld/(d-alpha), 1(Rd)) <= C parallel to D(alpha)u parallel to(L1(Rd; Rd)) for all u is an element of L-q(R-d) for some 1 <= q < d/(1 - alpha) such that D(alpha)u := VI(1-alpha)u is an element of L-1(R-d; R-d). We also give a counterexample which shows that in contrast to the case alpha = 1, the fractional gradient does not admit an L-1 trace inequality, i.e. parallel to D(alpha)u parallel to(L1(Rd; Rd)) cannot control the integral of u with respect to the Hausdorff content H-co(d-alpha). The main substance of this counter-example is a result of interest in its own right, that even a weak-type estimate for the Riesz transforms fails on the space L-1(H-co(d-beta)), beta is an element of [1, d). It is an open question whether this failure of a weak-type estimate for the Riesz transforms extends to beta is an element of (0, 1).